uczestnik wyscigu rowerowego porusza sie z szybkoscia 23km/h. Promien koła roweru wynosi 0,5m. Jaka jest liczba obrotow kola w jednej sekundzie
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V = 23 km/h = 23 *1000 m/3600 s ≈ 6,39 m/s
r = 0,5 m
t = 1 s
n = ?
V = 2πr / T ale T = t/n
V = 2πr /t/n
V = 2πrn/t
V * t = 2πrn
n = V *t /2πr
n = 6,38 m/s *1 s /2 *3,14 * 0,5 m
n = 6,38 m /3,14 m
n = 2